Latest TCS Aptitude Question SOLUTION: The sequence {A(n)} is defined by A1 = 2 and A(n+1) = A(n +2n) what is the value of A(100)

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The sequence {A(n)} is defined by A1 = 2 and A(n+1) = A(n +2n) what is the value of A(100)

Asked In TCS RADHIKA SHARMA (10 years ago)

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Read Solution (8) :

Aditya  (10 years ago)

this can also be written as :
A(n+1)=n^2+n+2
A(0+1)=0^2+0+2=2
A(1+1)=1^2+1+2=4
A(2+1)=2^2+2+2=8
A(3+1)=3^2+3+2=14
A(4+1)=4^2+4+2=22
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A(99+1)=99^2+99+2=9801+99+2=9902
hence,A(100)=9902

Is this solution Helpfull? Yes (14) | No (1)  

Kanshul Maharifa.M  (10 years ago)

A1=2
An+1=An+2n
A2=A1+2*1=4
A3=A2+2*2=8
A4=A3+2*3=14
A5=A4+2*4=22
...A99
Now solve the above eqns
A1+A2+A3+A4+...+A99+A100=A1+A2+A3+...+A99+2(1+2+3+4+5+...+99)
Now eliminate right hand side and left hand side upto A99 then,
A100=2(1+2+3+4+5+...+99)
using n integer formula[n*(n+1)/2]
n=99
99*(99+1)/2=99*100/2=9900/2=4950
Thus solution of A100=2*4950=9900

Is this solution Helpfull? Yes (7) | No (2)  

zeba salma  (10 years ago)

given A1=2 and An+1=An+2n and A100=?
lets find out A2=A1+1=A1+2=2+2=4
SIMILARLY A3=A2+1=A2+2*2=4+4=8
series is getting followed as 2,4,8...2^n
hence A100=A99+1=2^99

Is this solution Helpfull? Yes (6) | No (9)  

sujeet kumar keshri  (10 years ago)

given A(1)=2
A(n+1)=A(n)+2n
n=1, A(2)=(1*0+2)+2*1=4
n=2, A(3)=(2*1+2)+2*2=8
n=3, A(4)=(3*2+2)+2*3=14
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n=99, A(100)=(99*98+2)+2*99=9902

Is this solution Helpfull? Yes (4) | No   

PM  (10 years ago)

A 1 = A 0 = 2
so A 2 = A 3
and A 3 = A 6
A 4 = A 9
A 5 = A 12
so A 100 = A 297
now A 3 - A 2 = A 6 - A 3 ~ A 3
so A 1 - A 0 = 2 ~A 3
clearly A 100 = 2*100 = 200 Ans

Is this solution Helpfull? Yes (2) | No (6)  

M.Abhi Varsha  (10 years ago)

But zeba, if u calculate a4 u will get 14 which is not in d series...

Is this solution Helpfull? Yes (1) | No (1)  

naga sarath chand  (10 years ago)

just simple,see
a_n+1 = a_n + 2n,
so, a_5 = a_4+2*4 = a_3 +2*3+2*4(since again a_4=a_3 +2*3) = a_2 +2*2+2*3+2*4
= a_1 + 2(1+2+3+4),
so from this,
a_100 = a_1 + 2*(1+2+3+.....+99)= 2+2*(this is sum of n terms) = 2+2*(99*100/2)

= 2+9900 = 9902.

Is this solution Helpfull? Yes (1) | No   

Sneha Suman  (10 years ago)

Given- An+1=An+2n.
So An=An-1 + 2(n-1) and An-1=An-2 + 2(n-2).
We have An+1=An-2 + 2(n+(n-1)+(n-2)).
Taking a look at the pattern we can say that similarly -
An+1= A1 + 2(n+(n-1)+(n-2)......+1).
An+1=A1 + 2(n)(n+1)/2
An+1=2 + n(n+1).
Here n+1=100, n=99,
So A100= 2 + 99*100 = 9902(ANS)

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